python求亲密数对

Question

键盘输入N，N在2至2000之间，求2至N中的亲密数对。
就是A的（去掉1和自己的）因子和等于B，B的（去掉1和自己）因子和等于A，且A≠B。
如48和75是亲密数对。
48的因子和为2+3+4+6+8+12+16+24=75，
而75的因子和为3+5+15+25=48。
输入
200
输出
48 75
75 48
140 195
195 140
下面的程序对了一半

def sum_of_factors(num):
    total = 0
    for i in range(2, int(num/2) + 1):
        if num % i == 0:
            total += i
    return total
def find_amicable_numbers(n):
    result = []
    for i in range(2, n+1):
        a = i
        b = sum_of_factors(i)
        if a != b and sum_of_factors(b) == a:
            result.append((a, b))
    return result
def main():
    n = int(input())
    amicable_numbers = find_amicable_numbers(n)
    for pair in amicable_numbers:
        print(pair[0], pair[1])
 
if __name__ == "__main__":
    main()

Answer 1

你只考虑了b的因子和等于a，而没有考虑a的因子和等于b。
第12行改为：

if a != b and sum_of_factors(b) == a and sum_of_factors(a) == b:

完整代码：

def sum_of_factors(num):
    total = 0
    for i in range(2, int(num/2) + 1):
        if num % i == 0:
            total += i
    return total

def find_amicable_numbers(n):
    result = []
    for i in range(2, n+1):
        a = i
        b = sum_of_factors(i)
        if a != b and sum_of_factors(b) == a and sum_of_factors(a) == b: ####
            result.append((a, b))
    return result

def main():
    n = int(input())
    amicable_numbers = find_amicable_numbers(n)
    for pair in amicable_numbers:
        print(pair[0], pair[1])
 
if __name__ == "__main__":
    main()